TITLE: If there is already enough room to add all projections, does passing to matrices change anything?
QUESTION [8 upvotes]: Throughout, $A$ denotes a $*$-algebra. We always assume $A$ is representable in the sense that $A$ can be embedded into $B(H)$ for some Hilbert space $H$. The particular embedding is not important, and it is not assumed that the range of this embedding is closed, i.e. $A$ does need not be a $C^*$-algebra. 
Let $P(A)$ be the set of projections (self-adjoint idempotents) in $A$. Let $\sim$ denote the Murray-von Neumann equivalence relation on $P(A)$. That is, $e \sim e'$ when there exists $w \in A$ such that $w^*w = e$ and $ww^*=e'$.
Projections $e_1,e_2 \in A$ are called orthogonal if $e_1 e_2=0$. In this case, $e_2 e_1 = 0$ as well and $e_1 + e_2$ is another projection. That is, $P(A)$ is closed under orthogonal sums. The following fact states that addition of Murray-von Neumann equivalence classes is well-defined, whenever orthogonal representations are to be found.

Fact 1: Suppose that $e_1, e_2 \in P(A)$ are two orthogonal projections, and $e_1',e_2'$ are two more orthogonal projections. If $e_1 \sim e_1'$ and $e_2 \sim e_2'$, then $e_1 + e_2 \sim e_1' + e_2'$.
Proof:  Indeed, suppose $w_1,w_2 \in A$ have $w_1^*w_1 = e_1$, $w_1w_1^* = e_1'$ and $w_2^*w_2 = e_2, w_2w_2^* = e_2'$. One has $w_1^*w_2= (e_1'w_1)^* e_2'w_2 = w_1^* e_1' e_2' w_2 = 0$ and, similarly, $w_2^*w_1 = 0$. Using the latter two equalities, one sees $(w_1 + w_2)^*(w_1+w_2) = e_1 + e_2$. Similarly, $(w_1+w_2)(w_1+w_2)^* = e_1' + e_2'$ so that $(e_1+e_2) \sim (e_1' +e_2')$. 

As a simple corollary, we see that, if $A$ is spacious in the sense that, for each pair of projections $e_1,e_2 \in P(A)$, there exist $e_1',e_2' \in P(A)$ with $e_1 \sim e_1'$, $e_2 \sim e_2'$ such that $e_1'$ and $e_2'$ are orthogonal, then $P(A) / \sim$ is a commutative monoid with respect to orthogonal direct sum. 
Even if $A$ is not already spacious, then $M_\infty(A) = \bigcup_{n=1}^\infty M_n(A)$, which also acts faithfully on a Hilbert space, namely $\bigoplus_{n=1}^\infty H$, is spacious. This is follows easily from the fact that, if $e \in P(M_n(A))$, then $w = \left( \begin{smallmatrix} 0 & 0 \\ e & 0 \\ \end{smallmatrix} \right) \in M_{2n}(A)$ has $w^*w = \left( \begin{smallmatrix} e & 0 \\ 0 & 0 \\ \end{smallmatrix} \right)$, $ww^* = \left( \begin{smallmatrix} 0 & 0 \\ 0 & e \\ \end{smallmatrix} \right)$. We remark that:

$A \mapsto M_\infty(A)$ is a functor from representable $*$-algebras and $*$-homomorphisms, to the subcategory of representable $*$-algebras that are spacious.
$A \mapsto P(A) / \sim$ is a functor from spacious, representable $*$-algebras to commutative monoids. 

My question is the following:

Question: If $A$ is already spacious so that $P(A)/\sim$ is monoid without passing to matrices, does the corner inclusion $A \to M_\infty(A)$ induce a monoid isomorphism $P(A) / \sim \to P(M_\infty(A)) / \sim$? In other words: 

Is every $e \in P(M_\infty(A))$ Murray-von Neumann equivalent in $M_\infty(A)$ to an $e' \in A \subset M_\infty(A)$?
If $e,e ' \in A$ are Murray-von Neumann equivalent in $M_\infty(A)$, are they already in $A$?

REPLY [3 votes]: Inspired by George Lowther's solution, here is another more elementary example of a (nonunital) C*-algebra $A$ such that $A$ has no nonzero projections, but $M_2(A)$ does have nonzero projections.

Let $D$ be the sub algebra of $M_2(\mathbb{C})$ consisting of scalar multiples of the identity.
Let $C$ be the sub algebra of $M_2(\mathbb{C})$ consisting of matrices of the form $\begin{bmatrix} * & 0 \\ 0 & 0 \end{bmatrix}$. 
Let $A$ be the sub algebra of $C([0,1],M_2(\mathbb{C}))$ consisting of all paths $f$ for which $f(0)\in D$ and $f(1)\in C$. 

I claim there are no nonzero projections in $A$. To see this, note that a projection in $C([0,1],M_2(\mathbb{C}))$ is just a path of projections in $M_2(\mathbb{C}$), and the rank (which equals the trace) must stay constant along such a path. Since the only nonzero projection in $D$ has rank 2, while the only nonzero projection in $C$ has rank 1, there cannot be a nonzero projection in $A$. 
On the other hand, $M_2(A)$ has nonzero projections. Take, for example:
$$P(t) = U(t)
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0& 0 & 0 & 0 \\
0& 0 & 0 & 0 \\
\end{bmatrix} U(t)^{-1}$$
where $U : [0,1] \to M_4(\mathbb{C})$ is a path of unitaries with 
\begin{align*} U(0)=I_4 && U(1)=
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0& 1 & 0 & 0 \\
0& 0 & 0 & 1 \\
\end{bmatrix}
\end{align*}